Complete question:
The growth of a city is described by the population function p(t) = P0e^kt where P0 is the initial population of the city, t is the time in years, and k is a constant. If the population of the city atis 19,000 and the population of the city atis 23,000, which is the nearest approximation to the population of the city at
Answer:
27,800
Step-by-step explanation:
We need to obtain the initial population(P0) and constant value (k)
Population function : p(t) = P0e^kt
At t = 0, population = 19,000
19,000 = P0e^(k*0)
19,000 = P0 * e^0
19000 = P0 * 1
19000 = P0
Hence, initial population = 19,000
At t = 3; population = 23,000
23,000 = 19000e^(k*3)
23000 = 19000 * e^3k
e^3k = 23000/ 19000
e^3k = 1.2105263
Take the ln
3k = ln(1.2105263)
k = 0.1910552 / 3
k = 0.0636850
At t = 6
p(t) = P0e^kt
p(6) = 19000 * e^(0.0636850 * 6)
P(6) = 19000 * e^0.3821104
P(6) = 19000 * 1.4653739
P(6) = 27842.104
27,800 ( nearest whole number)
14^15/14^5=14^(15-5)=14^10
Hope this kinda helps !!
F(x) = 3x + 1
let f(x) = y
y = 3x + 1
y - 1 = 3x
3x = y - 1
x = (y - 1)/3
From y = f(x), x = f⁻¹(y)
<span>x = (y - 1)/3
</span>
f⁻¹(y) = (y - 1)/3
f⁻¹(7) = <span>(7 - 1)/3 = 6/3 = 2
</span>f⁻¹(7)<span> = 2</span>
Triangular sequence = n(n + 1)/2
If 630 is a triangular number, then:
n(n + 1)/2 = 630
Then n should be a positive whole number if 630 is a triangular number.
n(n + 1)/2 = 630
n(n + 1) = 2*630
n(n + 1) = 1260
n² + n = 1260
n² + n - 1260 = 0
By trial an error note that 1260 = 35 * 36
n² + n - 1260 = 0
Replace n with 36n - 35n
n² + 36n - 35n - 1260 = 0
n(n + 36) - 35(n + 36) = 0
(n + 36)(n - 35) = 0
n + 36 = 0 or n - 35 = 0
n = 0 - 36, or n = 0 + 35
n = -36, or 35
n can not be negative.
n = 35 is valid.
Since n is a positive whole number, that means 630 is a triangular number.
So the answer is True.
